Question

A cottage industry manufactures pedestal lamps and wooden shades, each requiring the use of a grinding/cutting machine and a sprayer. It takes $2h$ on grinding/cutting machine and $3h$ on the sprayer to manufacture a pedestal lamp. It takes $1h$ on the grinding/cutting machine and $2h$ on the sprayer to manufacture a shade.On any day, the sprayer is available for atmost $20h$ and the grinding/cutting machine for atmost $12h$. The profit from the sale of a lamp is $₹5$ and that from a shade is $₹3$. Assuming that the manufacturer can sell all the lamps and shades that he produces, the number of pedestal lamps and wooden shades to maximise the profit are respectively.

• A. 4,5
• B. 5,4
• C. 3,3
• D. 4,4

Answer 1

Answer: (D)

Let the manutacturer produces $x$ pedestal lamps and $y$ wouderi shades everyday. We consistruct lhe folluwing lable

Item	Number of packages	Time on grinding/cuttin g machine (inh)	Time onsprayer(in h)	Prófit (in ₹)
A	x	2x	3x	5x
B	y	y	2 y	3 y
Total	x+y	2x + y	3x + 2y	5 x + 3 y
Availability		12	20

The profits on a lamp is $₹5$ and on the shades $₹3$.
Our problem is to maximise $Z=5x+3y$...(i)
Subject to the constraints are $2x+y\le 12$...(ii)
$3x+2y\le 20$...(iii)
$x\ge 0,y\ge 0$...(iv)
Firstly, draw the graph of the line $2x+y=12$

x	0	6
y	12	0

Putting $(0,0)$ in the inequality $2x+y\le 12$, we have
$2\times 0+0\le 12\Rightarrow 0\le 12$(which is true)
So, the half plane is towards the origin.
Since, $x,y\ge 0$
So, the feasible region lies in the first quadrant.
Secondary, draw the graph of the line $3x+2y=20$

x	0	$\frac{20}{3}=6.6$
y	10	0

Putting $(0,0)$ in the inequality $3x+2y\le 20$, we have
$3\times 0+2\times 0\le 20$
$\Rightarrow 0\le 20$ (which is true)
So, the half plane is towards the origin.
On solving equations $2x+y=12$ and $3x+2y=20$, we get $B(4,4)$.
$\therefore$ Feasible region is OABCO.
The corner point of the feasible region are $O(0,0),A(6,0)$, $B(4,4)$ and $C(0,10)$. The values of $Z$ at these points are as follows

Corner point	$z=5x+3y$
$O(0,0)$	0
$A(6,0)$	30
$B(4,4)$	$32\to$ Maximum
$C(0,10)$	30

The maximum value of $Z$ is $₹32$ at $B(4,4)$.
Thus, the manufacturer should produce 4 pedestal lamps and 4 wooden shades to maximise his profits.